The number of electrons involved in the conversion of `MnO_(4)^(-)` to `MnO_(2)` is

Options

A. 3
B. 4
C. 1
D. 2

Explanation

To determine the number of electrons involved in the conversion of \( \text{MnO}_4^{-} \) to \( \text{MnO}_2 \), we will follow these steps:

### Step 1: Determine the oxidation state of manganese in \( \text{MnO}_4^{-} \)

The formula for calculating the oxidation state of manganese (Mn) in \( \text{MnO}_4^{-} \) is:
\[ \text{Oxidation state of Mn} + 4 \times \text{Oxidation state of O} = \text{Charge} \]
The oxidation state of oxygen (O) is typically \(-2\). Therefore:
\[ x + 4(-2) = -1 \]
\[ x - 8 = -1 \]
\[ x = +7 \]
So, the oxidation state of Mn in \( \text{MnO}_4^{-} \) is +7.

### Step 2: Determine the oxidation state of manganese in \( \text{MnO}_2 \)

Using the same formula for \( \text{MnO}_2 \):
\[ \text{Oxidation state of Mn} + 2 \times \text{Oxidation state of O} = 0 \]
\[ x + 2(-2) = 0 \]
\[ x - 4 = 0 \]
\[ x = +4 \]
So, the oxidation state of Mn in \( \text{MnO}_2 \) is +4.

### Step 3: Calculate the change in oxidation state

Now, we can see the change in oxidation state from \( \text{MnO}_4^{-} \) to \( \text{MnO}_2 \):
\[ \text{Change in oxidation state} = +7 \text{ (in MnO}_4^{-}) - +4 \text{ (in MnO}_2) = +3 \]
This indicates that manganese is reduced by 3 units.

### Step 4: Determine the number of electrons involved

Since the oxidation state decreases by 3, it means that manganese gains 3 electrons during the conversion from \( \text{MnO}_4^{-} \) to \( \text{MnO}_2 \).

### Conclusion

The number of electrons involved in the conversion of \( \text{MnO}_4^{-} \) to \( \text{MnO}_2 \) is **3 electrons**.

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